## 2016年11月17日木曜日

### 数学 - 2次曲線と直線 - 2次曲線 – 2次曲線と直線(楕円・双曲線と直線、接線の方程式)

$\begin{array}{l}双曲線とx軸の交点\left(a,0\right)、\left(-a,0\right)における接線の方程式は、それぞれ\\ x=a\\ ax={a}^{2}\\ \frac{ax}{{a}^{2}}-\frac{0y}{{b}^{2}}=1\\ \\ x=-a\\ -ax={a}^{2}\\ \frac{-ax}{{a}^{2}}-\frac{0y}{{b}^{2}}=1\end{array}$

$\begin{array}{l}x軸上にない点P\left({x}_{0},{y}_{0}\right)について。\\ y-{y}_{0}=m\left(x-{x}_{0}\right)\\ y=m\left(x-{x}_{0}\right)+{y}_{0}\\ を接線の方程式とする。\\ \frac{{x}^{2}}{{a}^{2}}-\frac{{\left(mx-\left(m{x}_{0}-{y}_{0}\right)\right)}^{2}}{{b}^{2}}=1\\ {b}^{2}{x}^{2}-{a}^{2}{\left(mx-\left(m{x}_{0}-{y}_{0}\right)\right)}^{2}={a}^{2}{b}^{2}\\ \left({b}^{2}-{a}^{2}{m}^{2}\right){x}^{2}+2{a}^{2}m\left(m{x}_{0}-{y}_{0}\right)x-{a}^{2}\left({\left(m{x}_{0}-{y}_{0}\right)}^{2}+{b}^{2}\right)=0\\ {x}_{0}=\frac{-{a}^{2}m\left(m{x}_{0}-{y}_{0}\right)}{{b}^{2}-{a}^{2}{m}^{2}}\\ {a}^{2}{x}_{0}{m}^{2}-{b}^{2}{x}_{0}={a}^{2}{x}_{0}{m}^{2}-{a}^{2}{y}_{0}m\\ m=\frac{{b}^{2}{x}_{0}}{{a}^{2}{y}_{0}}\\ y=\frac{{b}^{2}{x}_{0}}{{a}^{2}{y}_{0}}\left(x-{x}_{0}\right)+{y}_{0}\\ {a}^{2}{y}_{0}y={b}^{2}{x}_{0}x-{b}^{2}{x}_{0}{}^{2}+{a}^{2}{y}_{0}{}^{2}\\ {b}^{2}{x}_{0}x-{a}^{2}{y}_{0}y={b}^{2}{x}_{0}{}^{2}-{a}^{2}{y}_{0}{}^{2}\\ \frac{{x}_{0}x}{{a}^{2}}-\frac{{y}_{0}y}{{b}^{2}}=\frac{{x}_{0}{}^{2}}{{a}^{2}}-\frac{{y}_{0}{}^{2}}{{b}^{2}}\\ \frac{{x}_{0}x}{{a}^{2}}-\frac{{y}_{0}y}{{b}^{2}}=1\end{array}$

1. $\begin{array}{l}{x}^{2}+2{\left(-x+k\right)}^{2}=4\\ 3{x}^{2}-4kx+2{k}^{2}-4=0\\ \frac{D}{4}=4{k}^{2}-3\left(2{k}^{2}-4\right)\\ =-2{k}^{2}+12\\ =-2\left({k}^{2}-6\right)\\ 0個\left(|k|>\sqrt{6}\right)\\ 1個\left(|k|=\sqrt{6}\right)\\ 2個\left(|k|<\sqrt{6}\right)\end{array}$

2. $\begin{array}{l}{x}^{2}+2{\left(mx+2\right)}^{2}=4\\ \left(2{m}^{2}+1\right){x}^{2}+8mx+4=0\\ \frac{D}{4}=16{m}^{2}-4\left(2{m}^{2}+1\right)\\ =4\left(4{m}^{2}-\left(2{m}^{2}+1\right)\right)\\ =4\left(2{m}^{2}-1\right)\\ 0個\left(|m|<\frac{1}{\sqrt{2}}\right)\\ 1個\left(|m|=\frac{1}{\sqrt{2}}\right)\\ 2個\left(|m|>\frac{1}{\sqrt{2}}\right)\end{array}$